Talk:Pound-Star Notation
#*((2))*(1,3)*^*2**#*7 I want to solve #*((2))*(1,3)*^*2**#*7: n=1, #*((2))*(1,3)*^*2**#*7 n=7, #*((2))*(1,3)*^*2**# n=7, #*((2))*(1,3)*^*2**7 n=7, #*((2))*(1,3)*^*2*7*7*7*7*7*7*7 n=49, #*((2))*(1,3)*^*2*7*7*7*7*7*7 n=343, #*((2))*(1,3)*^*2*7*7*7*7*7 n=2401, #*((2))*(1,3)*^*2*7*7*7*7 n=16807, #*((2))*(1,3)*^*2*7*7*7 n=117649, #*((2))*(1,3)*^*2*7*7 n=823543, #*((2))*(1,3)*^*2*7 n=5764801, #*((2))*(1,3)*^*2 n=5764801, #*((2))*(1,3)*^25764801 (over 17 million characters long) n=5764801, #*((2))*(1,3)*^25764800*57648022 (over 23 million characters long) The next string would contain more than one trillion characters. -- 13:34, October 9, 2014 (UTC) Wrong. The seperator rules take precedence over the other rules. So you'd get n=7, #*((2))*(1,3)*^*2*#*#*#*#*#*#*#. -SJ224 14:39, October 9, 2014 (UTC) A second try: n=1, #*((2))*(1,3)*^*2**#*7 n=7, #*((2))*(1,3)*^*2**# n=7, #*((2))*(1,3)*^*2*#*#*#*#*#*#*# n=7, #*((2))*(1,3)*^*2*#*#*#*#*#*#*7 n=49, #*((2))*(1,3)*^*2*#*#*#*#*#*# n=49, #*((2))*(1,3)*^*2*#*#*#*#*#*49 n=2401, #*((2))*(1,3)*^*2*#*#*#*#*# n=2401, #*((2))*(1,3)*^*2*#*#*#*#*2401 n=5764801, #*((2))*(1,3)*^*2*#*#*#*# n=5764801, #*((2))*(1,3)*^*2*#*#*#*5764801 n=716, #*((2))*(1,3)*^*2*#*#*# n=716, #*((2))*(1,3)*^*2*#*#*716 n=732, #*((2))*(1,3)*^*2*#*# n=732, #*((2))*(1,3)*^*2*#*732 n=764, #*((2))*(1,3)*^*2*# n=764, #*((2))*(1,3)*^*2*764 n=7128, #*((2))*(1,3)*^*2 n=7128, #*((2))*(1,3)*^2(7128) (has over a googol characters) n=7128, #*((2))*(1,3)*^2(7128-1)*(7128+1)2 (has over a googol characters) The next string would contain more than one septuagintillion characters. -- 16:46, October 9, 2014 (UTC) HELP! Rulesets are too complicated for me. Help? -- A Large Number Googologist -- 15:54, October 12, 2014 (UTC) :what do you want us to do for you it's vel 16:03, October 12, 2014 (UTC) ::I'm really surprised that you can understand pretty much all of the R function, yet you cannot grasp the rules of this notation. LittlePeng9 (talk) 16:10, October 12, 2014 (UTC) @LP9 I just understand up to {0**} arrays. @vell make an easier ruleset to understand. (i'm still not generalized with some rulesets, especially where there are too long or make uses of symbols) Forget to add signature. -- A Large Number Googologist -- 17:06, October 12, 2014 (UTC) :what part specifically are you confused on it's vel 17:11, October 12, 2014 (UTC) EVERYTHING (not mad at you). -- A Large Number Googologist -- 17:21, October 12, 2014 (UTC) : Well, recently on chat you claimed that you understand all of R function up to Nested Array Notation, which is {S**...*} with any number of asterisks. LittlePeng9 (talk) 17:28, October 12, 2014 (UTC) Then i was wrong -- A Large Number Googologist -- 17:30, October 12, 2014 (UTC) I mean, i did not mean that i did understand up to {S***...***}, but up to {0**} -- A Large Number Googologist -- 17:31, October 12, 2014 (UTC) Soo, where is the easier explanation? -- A Large Number Googologist -- 17:50, October 12, 2014 (UTCi) : I find this definition to be quite easy if-then-else algorithm. I have no idea how to make it simplier. Try looking at the examples, maybe this will clarify some things. LittlePeng9 (talk) 17:56, October 12, 2014 (UTC) Still don't get Pound-Star Notation. -- A Large Number Googologist -- 18:07, October 12, 2014 (UTC) : I don't know how to help in that case. In my opinion, the definition cannot be made much more clean. Sorry. LittlePeng9 (talk) 18:09, October 12, 2014 (UTC) :( -- A Large Number Googologist -- 18:17, October 12, 2014 (UTC) Sorry, formating error. :( --~~~~ Just, forget it. -- A Large Number Googologist -- 18:18, October 12, 2014 (UTC) lmao this whole conversation it's vel 19:01, October 12, 2014 (UTC) Incorrect evaluation SuperJedi, Hyper-Exploding Pound-Star notation is comparable to f_{\omega^{\omega^\omega}} , not f_{\varepsilon_0} . Deedlit11 (talk) 00:03, October 17, 2014 (UTC) So I could have overestimated just about everything past Titanicol. That would be problematic. -SJ224 00:19, October 17, 2014 (UTC) Let's see: ^->w ^*->w+1 ^^->w*2 ^^^->w*3 {1}->w^2 {2}->w^3 {3}->w^4 {0:1}->w^w {1:1}->w^(w+1) {2:1}->w^(w+2) {0:2}->w^(w*2) {0:3}->w^(w*3) {0:0:1}->w^((w^2)+w) {1:0:1}->w^((w^2)+w+1) {0:1:1}->w^((w^2)+w*2) {0:0:2}->w^((w^2)*2) {0:0:3}->w^((w^2)*3) Yeah, I think you're right. :( -SJ224 00:28, October 17, 2014 (UTC) Missing definition of the question mark What does ? mean in your notation? n? -- A Large Number Googologist -- 21:59, October 17, 2014 (UTC) It meant 1 in an early draft, but I removed it from the definition. -SJ224 22:27, October 17, 2014 (UTC) Oh. -- A Large Number Googologist -- 22:40, October 17, 2014 (UTC) Proving that #*<<>> has level w^(w^w+w^2) in fgh @ = w^w^w <@> = w^(w^w+1) <<@>> = w^(w^w+2) 1 = w^(w^w+w) <1> = w^(w^w+w+1) 2 = w^(w^w+w*2) x = w^(w^w+w*x) As a result, #*<<>> has level w^(w^w+w^2) in fgh -- A Large Number Googologist -- 14:04, October 19, 2014 (UTC) Growth Rate of H#*<<>> The limit of previous notation was [] --> w^(w^w+w^2) * --> w^(w^w+w^2+w) ** --> w^(w^w+w^2+w*2) ^ --> w^(w^w+w^2*2) {1} --> w^(w^w+w^3) {1}{1} --> w^(w^w+w^3*2) {2} --> w^(w^w+w^4) {0:1} --> w^(w^w*2) {1:1} --> w^(w^(w+1)) {0:2} --> w^(w^(w*2)) {0:0:1} --> w^(w^(w^2)) @ --> w^(w^(w^w)) <@> --> w^(w^(w^w+1)) 1 --> w^(w^(w^w+w)) etc. As a result, H#*<<>> has limit e(0) -- A Large Number Googologist -- 19:05, October 20, 2014 (UTC) Oops! 1 --> w^(w^(w^w+w)), not just 1 (forget to add nowiki) -- A Large Number Googologist -- 19:08, October 20, 2014 (UTC) I knew it would almost certainly significantly exceed the lower bound I provided (w^w^(w*2)), but I didn't think it would exceed it by that much. -SJ224 10:15, October 21, 2014 (UTC) Yeah :D -- A Large Number Googologist -- 18:06, October 21, 2014 (UTC) Hello? Are you going to work on this? It has been like almost a month. -- From the googol and beyond -- 15:50, December 11, 2014 (UTC) Probably, eventually. -SJ224 11:45, December 15, 2014 (UTC) HELP Make Simpler Definition -- Notorious V.L. . 17:45, December 15, 2014 (UTC) I need Simpler Definition Please Explain Definition to Me -- Notorious V.L. . 17:46, December 15, 2014 (UTC) HELP Someone PLease Think for Me -- Notorious V.L. . 17:46, December 15, 2014 (UTC) :#* }^{1} = #* ..}}..1..}}..}}^{0}. Does That Help Cookiefonster (talk) 20:20, December 15, 2014 (UTC) HERE IS IT VELL you < your momma Happy now? OK IT WAS A JOKE If you have problems understanding this notation, ask SuperJedi, it may help. Or try looking again. Or understand this notation. -- From the googol and beyond -- 20:40, January 1, 2015 (UTC) A faster-growing revision I do this because there're long runs of steps with n being constant: #If there's only 1 entry, instead of doing multiplication, we use "square final number after "*" separator n times, then decrement n" (This is a dumping of n as its triangle into the final number). #If there're >1 numbers or the remaining separator isn't a single *, then increment n. #If the final entry in final parenthesized array is a 0, then increment n another time. That makes it faster. 15:29, February 27, 2018 (UTC) diction and operator clash fast -> swift: free:fast::swift:slow::quick:qualm::speedy:idle::hasty:laggy::fleet:laden. # is already used for the primorial. Can you find a new name? Alysdexia (talk) 19:28, April 28, 2018 (UTC)